Today, we will be bringing it all together. Given a function f (x), we know
how to find
• The domain of f
• The y-intercept of f
• The zeros of f
• The critical points of f
• Whether f is increasing/decreasing
• The inflection points of f
• Whether f is concave up or concave down
• The horizontal asymptotes of f
• The vertical asymptotes of f
This information is enough to give a fairly detailed picture of what a function
looks like.
For example, let’s say that I gave you the following information about a
function:
• f (x) is defined everywhere except x = 4.
• f (0) = −6
• f (x) = 0 when x = 2 and when x = 5
• f (x) has a critical point at x = −1 and at x = 7
• f (x) is increasing on (−∞, −1) and on (4, 7)
• f (x) is decreasing on (−1, 4) and (7, ∞).
• f (x) has inflection points at x = −3 and at x = 10.
• f (x) is concave up on (−∞, −3) and on (10, ∞), but concave down on
(−3, 4) and on (4, 10).
• f (x) has the unique horizontal asymptote of y = 0. f (x) satisfies limx→∞ f (x) =
0 and limx→−∞ f (x) = 0.
• f (x) has a vertical asymptote at x = 4.
Now, given this information, try to draw the graph of f (x).
Ideally, your graph should have had local maxima at x = −1 and at x = 7.
You should have noticed that limx→4− f (x) and limx→4+ f (x) were both −∞.
In other words, everyone should have drawn essentially the same graph.
The point here is that all of this information- the critical points, the inflection
points, the asymptotes- this is all stuff that you know how to find.
Now, let’s try out a practical example.
1
Problem 0.1. Sketch the graph of f (x), where f (x) =
x
x2 +5 .
Let’s do this systematically.
The domain of f (x) is (−∞, ∞). Notice that the denominator is always
strictly positive.
When x = 0, we have that f (x) = 0. So the graph passes through the point
(0, 0). This is the only place where f (x) = 0.
The derivative of f (x) is
f ′ (x) =
x2 + 5 − 2x2
(x2 + 5)2
which is
f ′ (x) =
5 − x2
.
(x2 + 5)2
√
√
We notice that f ′ (x) = 0 when x = 5 or when x = − 5. Therefore, these are
the critical points of f (x).
√
Plug in (for√example)
−10, 0, and 10 to √
see that f ′ (x) is negative on (−∞, − 5),
√
positive on (− 5, 5),
on√( 5, ∞). We
√ and negative
√
√ make our sign chart:
region x < − 5 − 5 < x < 5
x> 5
f ′ (x)
negative
positive
negative This shows that f (x)
f (x)
decreasing
increasing
decreasing
√
√
has a local minimum at x = − 5 and a local maximum at x = 5. We then
compute the second derivative:
f ′′ (x) =
(x2 + 5)2 (−2x) − (5 − x2 )(4x(x2 + 5)))
(x2 + 5)4
We can cancel out the x2 + 5 from the numerator and denominator:
f ′′ (x) =
(x2 + 5)(−2x) − 4x(5 − x2 )
(x2 + 5)3
We can also factor out a copy of −2x:
f ′′ (x) =
−2x((x2 + 5) + 2(5 − x2 ))
(x2 + 5)3
And now we simplify the numerator:
f ′′ (x) =
−2x(15 − x2 )
(x2 + 5)3
Again, the denominator is always
positive. The numerator is equal to zero
√
when x = 0 or when x = ± 15. √By plugging in points,
√ we can see that the
numerator is negative
for
x
<
−
15,
positive
for
−
15 < x < 0, negative
√
√
for 0 < x < 15, and positive for x > 15. Notice that this matches our
2
√
observation that
√ f (x) has a local minimum at x = − 5 and a local maximum at x = 5. (You should always do sanity checks like this in the middle
of a problem). Let’s
we know:
√ make
√ a complete√sign chart
√ to determine what √
√
√
region x < − 15 − 15 < x < − 5 − 5 < x < 0 0 < x < 5
5 < x < 15
f ′ (x)
negative
negative
positive
positive
negative
′′
f (x)
negative
positive
positive
negative
negative
f (x)
dec, CCD
dec, CCU,
inc, CCU
inc, CCD,
dec, CCD
Finally, it’s clear by the continuity of the function f that there are no vertical
asymptotes. We have that limx→∞ f (x) = limx→−∞ f (x) = 0, so y = 0 is the
horizontal asymptote.
As a final observation that’s useful for picturing the function, we notice that
f (−x) = −f (x), so f (x) is an odd function. This means that the graph should
be symmetric about the origin.
Now, try drawing the graph given all of the information above. This should
be enough to obtain a reasonable sketch. If you want even more detail, try
computing the value of f (x) at each of the critical points and inflection points.
3
√
x > 15
negative
positive
dec, CCU